Maths 5 Marks Questions

Given, the ratio of the volume of two spheres = 64 : 27

$\frac{\text{V}_1}{\text{V}_2}=\frac{64}{27}\Rightarrow\frac{\frac{4}{3}\pi\text{r}^3_1}{\frac{4}{3}\pi\text{r}^3_2{}}=\frac{64}{27}$

$\Rightarrow\ \ \ \Big(\frac{\text{r}_1}{\text{r}_2}\Big)^3=\Big(\frac{4}{3}\Big)^3\ \Big[\because\text{volume of sphere}=\frac{4}{3}\pi\text{r}^3\Big]$

$\Rightarrow\frac{\text{r}_1}{\text{r}_2}=\frac{4}{3}$

Let the surface areas of the two spheres be S₁ and S₂.

$\therefore\ \ \ \frac{\text{S}_1}{\text{S}_2}=\frac{4\pi\text{r}^2_1}{4\pi\text{r}^2_2}=\Big(\frac{\text{r}_1}{\text{r}_2}\Big)^2\ \ \ \ \Rightarrow\text{S}_1:\text{S}_2=\Big(\frac{4}{3}\Big)^2=\frac{16}{9}$

$\Rightarrow\ \ \ \text{S}_1:\text{S}_2=16:9$

Hence, the ratio of the their surface areas is 16 : 9.

According to the question,

Volume of cylider = Volume of a sphere

$\Rightarrow\ \ \pi\text{r}^2\text{h}=\frac{4}{3}\pi\text{r}^3 \ \ \ \Rightarrow\ \ \ \text{h}=\frac{4}{3}\text{r}$

$\because$ Diameter of the cylinder = 2r

$\therefore$ Increased diameter from hight of the cylinder $=2\text{r}-\frac{4\text{r}}{3}=\frac{2\text{r}}{3}$

Now, percentage increase in diameter of the cylinder $=\frac{\frac{2\text{r}}{3}\times100}{\frac{4}{3}\text{r}}50\%$

Hence, the diameter of the cylinder exceeds its height by 50%.

1. Given, radius of the base of a conical tent = 5m

And area needs to sit a student on the ground $=\frac{5}{7}\text{m}^2$

$\therefore$ Area of the base of a conical tent $=\pi\text{r}^2$

$=\frac{22}{7}\times5\times5\text{m}^2$

Now, $\text{number of student}=\frac{\text{Area of the base of a conical tent}}{\text{Area needs to sit a student on ground}}$

$=\frac{\frac{22\times5\times5\text{m}^2}{7}}{\frac{5}{7}}=\frac{22}{7}\times5\times5\times\frac{7}{5}=110$

Hence, 110 students can sit in the conical tent.

2. Given, area of the cloth to from a conical tent = 165m²

Radius of the base of a conical tent, r = 5m

Curved surface area of a conical tent = Area of cloth to from a conical tent

$\Rightarrow\pi\text{rl}=165$

$\Rightarrow\frac{22}{7}\times5\times\text{l}=165$

$\therefore \ \ \ \text{l}=\frac{165\times7}{22\times5}=\frac{33\times7}{22}=10.5\text{m}$

Now, height of a conical tent $=\sqrt{\text{l}^2-\text{r}^2}=\sqrt{(10.5)^2-(5)^2}$

$=\sqrt{110.25-25}=\sqrt{8525}=9.23\text{m}$

Volume of a cone (conical tent) $=\frac{1}{3}\pi\text{r}^2\text{h}=\frac{1}{3}\times\frac{22}{7}\times5\times5\times923$

$=\frac{1}{3}\times\frac{1550\times923}{7}=\frac{50765}{7\times3}=241.7\text{m}^3$

Hence, the volume of the cone (conical tent) is 241.7m³